One reason spectroscopists study excited states is to gain information about the energies of orbitals that are unoccupied in an
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Answer:
ΔG° = -80.9 KJ
Assuming this reaction takes place at room temperature (25 °C):
K=
Explanation:
1) Reduction potentials
First of all one should look up the reduction potentials for the species envolved:
→
E°red=1.61V
→
E°red=0.771V
2) Redox pair
Knowing their reduction pontentials one can determine a redox pair: one species must oxidate while the other is reducing. <u>Remember: the table gives us the reduction potential, so if we want to know the oxidation potential all that has to be done is reverce the equation and change the potencial signal (multiply to -1).</u>
1) reduces while
oxidates
(oxidation) →
E°oxi=-0.771V
(reduction) →
E°red=1.61V
(overall equation) →
E°=Ereduction + Eoxidation= 1.61 v+(-0.771 v) = 0.839v
The cell potential can also be calculated as the cathode potencial minus the anode potential:
E° = E cathode - E anode =1.61 v - 0.771 v=0.839 v
3) Gibbs free energy and Equilibrium constant
ΔG°=-nFE°, where 'n' is the number of electrons involved in the redox equation, in this case n is 1. 'F' is the Faraday constant, whtch is 96500 C. E° is the standard cell potencial.
ΔG°=-nFE°=-1*96500*0.839
ΔG° = - 80963 J = -80.9 KJ
The Nerst equation gives us the relation of chemical equilibrium and Electric potential.
E=E°-
Where 'R' is the molar gas constant (8.314 J/mol)
It's known that in the equilibrium E=0, so the Nerst equation, at equilibrium, becomes:
E°=
Isolating for 'K' gives:
This shows that 'K' is a fuction of temperature. Assuming this reaction takes place at room temperature (25 °C):
K=